Corresponding to distinct eigenvalues of a they form

This preview shows page 10 - 13 out of 17 pages.

corresponding to distinct eigenvalues of A, they form an eigenbasis for R2. Thus x0 = c1v1 + cTo compute c1 and c2 we apply row reduction to the augmented matrix [v1 v2 .This shows that c= 4, c2 = 3 and x0 = 4v1 3v2. Since v2 are eigenvectors corresponding tothe eigenvalues 5and 4 respectively, xk = Akx0 = Ak(4v1 3= 4Akv1 3Akv= 4(5)k v1 3(4)k v2 k 0. It follows that Axk = 4(5)k Av1 3(4)k = 4(5)k+1v1 3(4)k+1v2 = xkThis shows that {xk} solves the difference equation. Consequently, .015 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3,1, and 3 and corresponding eigenvectors v.If {xk} is the solution of the difference equation xk+1 = Axk, determine x1 3 3.x1 =11 correct11 4.x1 =11 3 5.x1 =111 1 6.x1 =11 3Explanation:To find x1 we must compute Ax0. First, we express express x0 in terms of v1,v2, and v3: c1v1 + c2v2 + c3vThis is certainly possible as the eigenvectors v1,v2, and v3 are linearly independent because the eigenvalues are distinct. Hence they form a basis for R3. The row reduction xk = 4(5)k v1 3(4)k
2v2
x
1
v1 and
v2)
2
for
Av2
+1.
x,
.
3
x0 =
3 .
v2
HW 10 gilbert 1 3 1 [v1 v2 v3 x0 ] =1 1 4 1 1 0 0 0 1 0 0 0 1 shows that x0 = 2v1 2v2 v3. But v1,v2 are eigenvectors for the respective eigenvalues 3,1 and 3, so x1 = Ax0 = A(2v1 2v2 = 2Av1 2Av2 A= 2 · (3)v1 2 · (1)v2 (3).Consequently, .016 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3, 2, and 3 and corresponding eigenvectors v.Determine the solution {xk} of the difference equation xk+1 = Axk, k k 1.xk = 2(3) v1 2(2) v2 (3) kk 2.xk = 2(3) v1 2(2) v2 + (kk 3.xk = 2(3) v1 2(2) v2 (kk 4.xk = 2(3) v1 + 2(2) v2 (3) kk 5.xk = 2(3) v1 + 2(2) v2 + (3) k k 6.xk = 2(3) v1 + 2(2) v2 + (correctExplanation:Since v1, v2, and v3 are eigenvectors corresponding to distinct eigenvalues of Athey form an eigenbasis for R3. Thus x0 c1v1 + c2v2 + c3To compute c1, c2, and c3 we apply row reduction to the augmented matrix 1 2 1 [v1 v2 v3 x0 ] =3 7 3 3 150 1 2 1 0 0 0 0 3 x1 =11 1 11 7 2 3 3 92 21and v3 v3) v3v2x.kv3k3) v3k3) v3kv3kv3k3) v3, = v35 4 16 30.2 1 1
HW 10 gilbert 12 This shows that c1 = 2, c2 = 2, c3 = 1 and 2v1+2v2+v3. Since v1, v2, and veigenvectors corresponding to eigenvalues 3, 2, 3 respectively, xk = Akx0 = Ak(2v1 + 2v2 + = 2Akv1 + 2Akv2 + Ak= 2(3)k v1 + 2(2)k v2 + (3)for k 0. It follows that Axk = 2(3)k Av1 + 2(2)k Av2 + (3)k 2(3)k+1v1 + 2(2)k+1v2 + (3)k+1v3 = xk
x0 =
3 are
the
v3)
v3
k v3
Av3 =
+1.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture